The integer solution set of inequality 2 ≤ X & # 178; - 2x < 8 is Quick answer

The integer solution set of inequality 2 ≤ X & # 178; - 2x < 8 is Quick answer

From 2 ≤ X & # 178; - 2x, we get: X & # 178; - 2x + 1 ≥ 3, and the solution is: X ≥ 1 + radical 3, or X ≤ 1-radical 3
From X & # 178; - 2x < 8: (x-4) (x + 2)
2≤x²-2x＜8
3≤(x-1)²＜9
The integer solution sets of inequality 2 ≤ X & # 178; - 2x < 8 are: - 1 and 3
If the line x + 2Y + M = 0 is tangent to the circle C: x2 + Y2 + 2x-4y = 0 after translation according to vector a = (- 1, - 2), then the value of real number m is equal to ()
A. 3 or 13b. 3 or - 13C. - 3 or 7d. - 3 or - 13
The line x + 2Y + M = 0 is translated into (x + 1) + 2 (y + 2) + M = 0 according to the vector a = (- 1, - 2), that is, x + 2Y + m + 5 = 0. Circle C: x2 + Y2 + 2x-4y = 0, that is, (x + 1) 2 + (Y-2) 2 = 5, which represents a circle with C (- 1,2) as the center and radius equal to 5. According to the tangent of the translated line and circle, the distance from the center of the circle to the straight line is equal to the radius, that is & nbsp; |− 1 + 4 + m + 5 | 5 = 5; M = - 3 or M = - 13, so D
The solution set of inequality 2x2-x-1 > 0 is______ .
Inequality 2x2-x-1 ＞ 0, factorization: (2x + 1) (x-1) ＞ 0, can be changed into: 2x + 1 ＞ 0 − 1 ＞ 0 or 2x + 1 ＜ 0 − 1 ＜ 0, solution: X ＞ 1 or X ＜ - 12, then the solution set of the original inequality is (− ∞, − 12) ∪ (1, + ∞). So the answer is: (− ∞, − 12) ∪ (1, + ∞)
If we translate the line x-2y + M = 0 according to the vector a = (- 1, - 2), the resulting line is tangent to the circle x2 + Y2 + 2x-4y =, then the value of M is?
Straight line x-2y + M = 0
That is y = x / 2 + m / 2
The line obtained by translating the line y = x / 2 + m / 2 according to the vector a = (- 1, - 2)
y=1/2*(x+1)-2+m/2
=1/2x-3/2+m/2
It is tangent to the circle x2 + Y2 + 2x-4y,
simultaneous
The discriminant is equal to zero
The solution is m = 3 or 13
For example: solve the X inequality X & # 178; - 2x + a ≥ 0 (urgent!
A:
x^2-2x+a>=0
(x-1)^2>=1-a
1) When 1-A = 1:
(x-1)^2>=0>=1-a
The solution of inequality is real number r
2) When 1-A > 0, a = √ (1-A) or X-1 = 1 + √ (1-A) or X
If the line 2x-y + C = 0 is tangent to the circle x square + y square = 5 after being translated according to the vector a = (1, - 1), the value of C is obtained
The x-axis moves 1 unit X-1 to the right and 1 unit x + 1 to the left
Y direction, up Y-1, down y + 1
Translation vector 1, - 1
So the line becomes
2(x-1)-(y+1)+c=0
2x-y+c-3=0
The future is very simple
Solution inequality: - X & # 178; - 2x + 8 ≤ 0
-x²-2x+8≤0
x²+2x-8≥0
(x-2)(x+4)≥0
X ≥ 2 or X ≤ - 4
Transformation: x ^ 2 + 2x-8 > = 0
Do factor score (x + 4) (X-2) > = 0
Get: x > = 2 or X
Given the circle C: (x-3) ^ 2 (y-4) ^ 2 = 4 and the line L: x + 2Y + 2 = 0, the lines m and N pass through the outer fixed point a (1,0) of the circle C. (1) if the line m is tangent to the circle C, find the equation of the line M. (2) if the line n intersects with the circle C at R and Q, intersects with the line L at n, and the midpoint of the line PQ is m, prove that am multiplied by am is the fixed value
Circle C: (x-3) ^ 2 + (y-4) ^ 2 = 4
From the circle formula, we can get the circle point (3,4) with radius 2. Let the equation of the line m be y = KX + B, that is, kx-y + B = 0. Because the line m passes through the fixed point a (1,0), it is substituted into the equation, that is, K + B = 0. Because the line m is tangent to the circle C, the distance between the circle point and the line is the radius of the circle C
1. Solve the inequality 2x & # 178; + kx-k ≤ 0 about X
2. Let f (x) = - 2x-2ax + A + 1, where - 1 ≤ x ≤ 0, a ≥ 0, and the maximum value of F (x) is d
(1) Let a denote D
(2) Find the minimum value of D and point out the value of a at this time
1.(2x-1)(x+1)
The first question is a typical classification discussion. You can go on carefully. The place of the classification is good. Don't forget that the discriminant is less than zero
1. Classification discussion, see discriminant, three cases.
2, (1) f (x) increases monotonically at [- 1,0], so f (0) = D, i.e
a+1=d
So a + D = 1
(2) Because d = a + 1 is a function of degree a, and a ≥ 0
So D ≥ 1
So the minimum value of D is 1, and the value of a is 0
If the moving point (x, y) on the plane moves on X + 2Y = 3, find the minimum value of x square of 2 + y square of 4
When x = 1, y = 1, the minimum value is 6